Find the missing frequencies in the following frequency distribution whose mean is 34.
$\begin{matrix} x & 10 & 20 & 30 & 40 & 50 & 60 & Total f & 4 & f_{1} & 8 & f_{2} & 3 & 4 & 35 \end{matrix}$

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Solution

We know that,
Mean = $\frac{\sum x i f i}{\sum f i}$

Mean = $\frac{(10 \times 4 + 20 \times f 1 + 30 \times 8 + 40 \times f 2 + 50 \times 3 + 60 \times 4)}{35}$
⇒ $34 = \frac{(40 + 20 f_{1} + 240 + 40 f_{2} + 150 + 240)}{35}$
⇒ $34 \times 35 = 670 + 20 f_{1} + 40 f_{2}$

⇒ $1190 - 670 = 20 f_{1} + 40 f_{2}$
⇒ $20 f_{1} + 40 f_{2} = 520$
⇒ $20 (f_{1} + 2 f_{2}) = 520$
⇒ $f_{1} + 2 f_{2} = \frac{520}{20}$

⇒ $f_{1} + 2 f_{2} = 26$

⇒ $f_{1} = 26 - 2 f_{2}$ ......................(1)

Also, $4 + f_{1} + 8 + f_{2} + 3 + 4 = 35$
⇒ $19 + f_{1} + f_{2} = 35$
⇒ $f_{1} + f_{2} = 35 - 19$
⇒ $f_{1} + f_{2} = 16$
⇒ $26 - 2 f_{2} + f_{2} = 16$
(from (1))
⇒ $26 - f_{2} = 16$
⇒ $26 - 16 = f_{2}$
⇒ $f_{2} = 10$
Putting the value of $f_{2}$ in (1), we get

$f_{1} = 26 - 2 (10) = 6$
Hence, the value of $f_{1}$ and $f_{2}$ is 6 and 10, respectively.

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Find the missing frequencies in the following frequency distribution whose mean is 34. x102030405060Totalf4f18f23435

Find the missing frequencies in the following frequency distribution whose mean is 34.
$\begin{matrix} x & 10 & 20 & 30 & 40 & 50 & 60 & Total f & 4 & f_{1} & 8 & f_{2} & 3 & 4 & 35 \end{matrix}$